Question Number 132600 by Ahmed1hamouda last updated on 15/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 132597 by Study last updated on 15/Feb/21 Commented by mr W last updated on 15/Feb/21 $$\frac{\mathrm{156}}{{x}+\mathrm{7}}=\frac{\mathrm{156}}{{x}}−\mathrm{14} \\ $$$$\mathrm{156}{x}=\mathrm{156}\left({x}+\mathrm{7}\right)−\mathrm{14}{x}\left({x}+\mathrm{7}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{78}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left({x}+\mathrm{13}\right)=\mathrm{0}…
Question Number 132598 by Ahmed1hamouda last updated on 15/Feb/21 $$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{voloume}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{z}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{y}+\mathrm{z}=\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 67058 by mhmd last updated on 22/Aug/19 $${find}\:{the}\:{area}\:{abounded}\:{y}=\sqrt{{x}−\mathrm{2}} \\ $$$${and}\:{y}={x}−\mathrm{2}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 67059 by mhmd last updated on 22/Aug/19 $${find}\:{the}\:{area}\:{abounded}\:{y}=\sqrt{{x}−\mathrm{2}} \\ $$$${and}\:{y}={x}−\mathrm{2}\:? \\ $$ Commented by Tony Lin last updated on 22/Aug/19 $${when}\:{y}=\sqrt{{x}−\mathrm{2}}={x}−\mathrm{2} \\ $$$${x}−\mathrm{2}=\left({x}−\mathrm{2}\right)^{\mathrm{2}}…
Question Number 132594 by Study last updated on 15/Feb/21 Commented by mr W last updated on 15/Feb/21 $${the}\:{sum}\:{of}\:{two}\:{complex}\:{numbers} \\ $$$${can}\:{be}\:{a}\:{real}\:{number},\:{e}.{g}.\: \\ $$$$\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}−{i}\right)=\mathrm{2} \\ $$$${the}\:{product}\:{of}\:{two}\:{complex}\:{numbers} \\…
Question Number 67057 by Tony Lin last updated on 22/Aug/19 $$\left(\mathrm{1}\right){find}\:\cap_{{n}=\mathrm{1}} ^{\infty} \left[\mathrm{0},\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\left(\mathrm{2}\right){find}\:\cup_{{n}=\mathrm{2}} ^{\infty} \left[\frac{\mathrm{1}}{{n}},\:\mathrm{1}−\frac{\mathrm{1}}{{n}}\right] \\ $$ Answered by Kunal12588 last updated on…
Question Number 67055 by Tony Lin last updated on 22/Aug/19 $${let}\:\mathbb{Z}_{+} =\mathbb{N}\cup\left\{\mathrm{0}\right\},\:{f}:\:\mathbb{Z}_{+} ×\mathbb{Z}_{+} \rightarrow\mathbb{Z}_{+} \\ $$$${f}\left({m},\:{n}\right)=\frac{\left({m}+{n}\right)\left({m}+{n}+\mathrm{1}\right)}{\mathrm{2}}+{m} \\ $$$${prove}\:{that}\:{f}\:{is}\:{a}\:{one}-{to}-{one}\:{function} \\ $$$${and}\:{also}\:{an}\:{onto}\:{function} \\ $$ Terms of Service…
Question Number 67038 by mathmax by abdo last updated on 22/Aug/19 $${calculate}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{1}+\mathrm{2}^{{sinx}} }{dx}\:\:\:{with}\:{n}\:{integr}. \\ $$ Commented by mathmax by abdo last updated…
Question Number 132575 by nadovic last updated on 15/Feb/21 $$\mathrm{Hello}\:\mathrm{Tinku}\:\mathrm{Tara},\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{new}\:\mathrm{phone} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{am}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{remember}\:\mathrm{the}\: \\ $$$$\mathrm{password}\:\mathrm{for}\:\mathrm{my}\:\mathrm{previous}\:\mathrm{account}.\:{Is} \\ $$$${there}\:{anyway}\:{to}\:{help}\:{me}\:{retrieve}\:{it}? \\ $$ Commented by mr W last updated on…